Rewrite 18y x 4x using Commutative Property

15x+3x+19= -17 please help me find this answer

masha68 [24]

Answer:

${ \rm{15x + 3x + 19 = - 17}} \\ \\ { \rm{18x + 19 = - 17}} \\ \\ { \rm{18x = - 17 - 19}} \\ \\ { \rm{18x = - 36}} \\ \\ { \boxed{ \rm{x = - 2}}}$

5 0

2 years ago

Read 2 more answers

2x - 4 > 4 ...................

Veseljchak [2.6K]

Answer:

x>4

Step-by-step explanation:

8 0

3 years ago

Find all real solutions to the equation (x² − 6x +3)(2x² − 4x − 7) = 0.

Jet001 [13]

Answer:

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

Step-by-step explanation:

Relation given in the question:

(x² − 6x +3)(2x² − 4x − 7) = 0

Now,

for the above relation to be true the following condition must be followed:

Either (x² − 6x +3) = 0 ............(1)

or

(2x² − 4x − 7) = 0 ..........(2)

now considering the equation (1)

(x² − 6x +3) = 0

the roots can be found out as:

$x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

for the equation ax² + bx + c = 0

thus,

the roots are

$x = \frac{-(-6)\pm\sqrt{(-6)^2-4\times1\times(3)}}{2\times(1)}$

or

$x = \frac{6\pm\sqrt{36-12}}{2}$

or

$x = \frac{6+\sqrt{24}}{2}$ and, x = $x = \frac{6-\sqrt{24}}{2}$

or

$x = \frac{6+2\sqrt{6}}{2}$ and, x = $x = \frac{6-2\sqrt{6}}{2}$

or

x = 3 + √6 and x = 3 - √6

similarly for (2x² − 4x − 7) = 0.

we have

the roots are

$x = \frac{-(-4)\pm\sqrt{(-4)^2-4\times2\times(-7)}}{2\times(2)}$

or

$x = \frac{4\pm\sqrt{16+56}}{4}$

or

$x = \frac{4+\sqrt{72}}{4}$ and, x = $x = \frac{4-\sqrt{72}}{4}$

or

$x = \frac{4+\sqrt{2^2\times3^2\times2}}{2}$ and, x = $x = \frac{4-\sqrt{2^2\times3^2\times2}}{4}$

or

$x = \frac{4+(2\times3)\sqrt{2}}{2}$ and, x = $x = \frac{4-(2\times3)\sqrt{2}}{4}$

or

$x = \frac{2+3\sqrt{2}}{2}$ and, $x = \frac{2-(3)\sqrt{2}}{2}$

Hence, the possible roots are

x = 3 + √6 ; x = 3 - √6 ; $x = \frac{2+3\sqrt{2}}{2}$ ; $x = \frac{2-(3)\sqrt{2}}{2}$

7 0

2 years ago

HVAC technician average salary = $28/hour with a one-year Tech school certification costing an average of $7,500. What would you

Juli2301 [7.4K]

Answer:

56,000

Step-by-step explanation:

6 0

2 years ago

2 Here are two equations:

MatroZZZ [7]

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Step-by-step explanation:

a. Decide whether each neither of the equations,

i (3,4)

ii. (4,2.5)

ill. (5,5)

iv. (3,2)

To decide whether each point is solution to equations or not we will put the point in the equations

Equations are:

Equation 1: 6x + 4y = 34

Equation 2: 5x – 2y = 15

i (3,4)

Putting in Equation 1:

$6(3) + 4(4) = 34\\18+16=34\\34=34\\$

Putting in Equation 2:

$5(3) - 2(4) = 15\\15-8 = 15\\7\neq 15$

ii. (4,2.5)

Putting in Equation 1:

$6(4) + 4(2.5) = 34\\24+10=34\\34=34\\$

Putting in Equation 2:

$5(4) - 2(2.5) = 15\\20-5 = 15\\15=15$

ill. (5,5)

$6(5) + 4(5) = 34\\30+20=34\\50\neq 34$

Putting in Equation 2:

$5(5) - 2(5) = 15\\25-10 = 15\\15=15$

iv. (3,2)

$6(3) + 4(2) = 34\\18+8=34\\26\neq 34$

Putting in Equation 2:

$5(3) - 2(2) = 15\\15-4 = 15\\11\neq 15$

Hence,

(3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. Is it possible to have more than one (x, y) pair that is a solution to both

equations?

The simultaneous linear equations' solution is the point on which the lines intersect. Two lines can intersect only on one point. So a linear system cannot have more than one point as a solution

So,

a. (3,4) is only the solution to Equation 1.

(4, 2.5) is the solution to both equations

(5,5) is the solution to Equation 2

(3,2) is not the solution to any equation.

b. No, it is not possible to have more than one (x,y) pair that is solution to both equations

Keywords: Linear equations, Ordered pairs

Learn more about linear equations at:

#LearnwithBrainly

8 0

3 years ago